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CHEM 433 – 10/6/11

CHEM 433 – 10/6/11. III. 1st Law: Thermochemistry (2.7-2.9) - finish D f H°’s - comment about D r U° - T dependence of  H READ: FINISH CHAPTER 2 HW#4: DUE tomorrow … see email about #2.10.

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CHEM 433 – 10/6/11

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  1. CHEM 433 – 10/6/11 III. 1st Law: Thermochemistry (2.7-2.9) - finish DfH°’s - comment about DrU° - T dependence of H READ: FINISH CHAPTER 2 HW#4: DUE tomorrow … see email about #2.10

  2. Standard Enthalpies of Formation: rH° to produce 1 mol of a given substance from elements in their reference states (most stable from, 1 bar, some specified T - often 298K )

  3. Using “heats” of Formation : rH° for a given reaction can be obtained via: {The sum of rH°’s for the products} – {The sum of rH°’s for the reactants}

  4. Lets get DcH° for C6H6 from these data: DfH°(C6H6)= + 49.0 kJ/mol DfH°(O2 (g))= ? (hello …) DfH°(CO2)= – 393.5 kJ/mol DfH°(H2O)= – 285.8 kJ/mol

  5. Using DfH°’s is just Hess’ Law in disguise! (see picture on previous slide – 2 back …) Lets get DcH° for C6H6 from these reactions: 6 C (s, gr) + 3 H2(g) —> C6H6 DH° = DfH° O2 (g) —> O2 (g) DH° = DfH° C (s, gr) + O2(g) —> CO2 DH° = DfH° H2 (g) + ½ O2 (g) —> H2O (l) DH° = DfH°

  6. Units on b must be K-1 ! ? ! ?

  7. T dependence of H, note that:

  8. Calculate rH° at 373K for: H2 (g) + 1/2 O2 —> H2O (g) H298K = -241.8 kJ Cp,m: 28.82 29.36 33.6 (in J/Kmol)

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